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3 years ago

Which statement accurately describes nutritional dietary supplements?

Physics

1 answer:

vladimir2022 [97]3 years ago

7 0

Answer:

its b

Explanation:

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A massless rod of length L has a small mass m fastened at its center and another mass m fastened at one end. On the opposite end

konstantin123 [22]

Answer:

onservation of energy

U top = K bottom

(m + m)*g*L = 1/2*I*?^2 where I = m*(L/2)^2 + m*L^2 = 1.25*m*L^2

So 2m*g*L = 1/2*1.25*m*L^2*?^2

So ? = sqrt(3*g*/(1.25*L) ) = sqrt(12g/5L)

3 0

3 years ago

Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter

goblinko [34]

Answer:

134.77 mm

Explanation:

Wave length of light λ = 599 x 10⁻⁹ m

Slit separation d = 20 x 10⁻⁶ m

Screen distance D = 3 m

Distance of second dark fringe from centre

= 1.5 x λ D / d

Putting the values given above

distance = $\frac{1.5\times599\times10^{-9}\times 3}{20\times10^{-6}}$

= 134.77 x 10⁻³ m

= 134.77 mm.

7 0

3 years ago

What type of galaxy is shown?

dexar [7]

Elliptical, because the shape of the galaxy isn’t like the others. It is unique to its own and doesn’t have another to compare to

8 0

3 years ago

Read 2 more answers

The temperature scale which starts at absolute zero is the _____.

oksian1 [2.3K]

The temperature scale which starts at absolute zero is the Kelvin scale. The correct option in respect to the given question is the last option. William Thompson was the British scientist and inventor that invented the Kelvin scale. William Thompson was also popularly known as Lord Kelvin.His discovery of the Kelvin scale is considered one among the three best scales in use for measuring temperatures.Each measuring unit of this scale is never called a degree but a Kelvin. This specialized scale gives the option of measuring temperature in both centigrade and Fahrenheit.

6 0

3 years ago

Read 2 more answers

Consider steady-state conditions for one-dimensional conduction in a plane wall having a thermal conductivity k = 50 W/m · K and

tatuchka [14]

Answer:

solution:

dT/dx =T2-T1/L

q_x = -k*(dT/dx)

dT/dx= (-20-50)/0.35==> -280 K/m

q_x =-50*(-280)*10^3==>14 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (2)

dT/dx= (-10+30)/0.35==> 80 K/m

q_x =-50*(80)*10^3==>-4 kW

Case (3)

q_x =-50*(160)*10^3==>-8 kW

T2=T1+dT/dx*L=70+160*0.25==> 110° C

Case (4)

q_x =-50*(-80)*10^3==>4 kW

T1=T2-dT/dx*L=40+80*0.25==> 60° C

Case (5)

q_x =-50*(200)*10^3==>-10 kW

T1=T2-dT/dx*L=30-200*0.25==> -20° C

note:

all graph are attached

6 0

3 years ago